# Why does force x distance not equal kinetic energy for objects rolling down an inclined plane?

Imagine two objects at rest at the top of an inclined plane which are capable of rolling and which are identical in every way (mass, radius, etc) with the exception of how their mass is distributed. One is focused towards the centre and the other is focused towards the outside thus giving them a different moment of inertia.

Both will have the same force of gravity pulling them down the incline and it will act through the same distance from the top to the bottom. The same goes for the torque (due to static friction) and the angular distance.

Thus it seems the work done on each is the same in terms of both rotational and translational kinetic energy (KE) and the sum of each for each object will be mgh.

That would normally lead me to conclude that they would have the same speeds but in this situation that clearly isn't true because they don't both accelerate at the same rate due to the difference in their moment of inertia. I can see that the object with the higher moment of inertia will have more energy in terms of its rotational KE and less in terms of translational KE regardless of the fact that they have both had the same force acting through the same distance, and torque through angular distance.

There seems to be a disconnect here and I'm finding it very confusing. Is the work done by a force through a distance not always equal to the objects kinetic energy as it appears here or am I just missing the obvious again?

• In the case of the rolling object you have to consider the work done on each individual volume element of it, because $\vec F\cdot d\vec s$ varies with time and position. – my2cts Sep 13 '18 at 18:25

The work done by force through a distance is the change in kinetic energy. However the kinetic energy here has two components: one due to the translational movement of the objects, one due to the fact that they are spinning. You must take both into account, you can't just take the translational component and ignore the other one.

Here is an extreme example of this. Consider two objects, one of which is a puck or something on a smooth surface (like ice or something with a coefficient of friction low enough that it can be ignored), and the other is a disk with an axle which is firmly fixed to some very large object, like the Earth, but which is free to rotate about its axle.

First of all we attach a light string to the first object and pull it with a given force through a given distance, parallel with the surface. At the end of that time the object has a bunch of kinetic energy, and all of it is in the translational motion of the object. And we can compute its final velocity ($m v^2/2 = Fd$ where $F$ is the force and $d$ is how far we pulled it) and so on.

Now we wind a similar light string around the disk, and pull that string through the same distance with the same force (and imagine that the string conveniently comes detached at the end of the pull in both cases). At the end of this process the disk has no translational KE at all, because its axle is fixed to the Earth. But it's spinning furiously, because what I've described here is essentially a child's spinning top or something very like it.

So the work I've done on both of these objects is the same, and so they have the same KE at the end of it (neglecting friction). And in both cases the energy was added by moving the same force through the same distance (pulling on the strings in other words). But they have it in quite different forms.

• To be clear: work done by force through a distance is only part of the contributions to kinetic energy. One must also add the work done by torque through an angular displacement. – garyp Sep 13 '18 at 11:27
• I think I'm just having a hard time getting my head around the fact that two objects can have the same net force acting through the same distance and yet their translational KE is different, and the same goes for the net torque through the same angular distance resulting in different rotational KE. – An_African_Ape Sep 13 '18 at 11:31
• I've added what I hope is an informative example by taking the extreme case where one of the objects ends up with no translational KE at all. – tfb Sep 13 '18 at 12:06

The moment inertia about its centre of mass of an object of mass $m$ and radius $r$ is $kmr^2$.

The value of $k$ ranges from $k=0$ for a point mass at the centre of a massless wheel of radius $r$ to $k=1$ where the mass is concentrated at a distance $r$ from the centre of mass.

If there is no slipping the relationship between the linear speed $v$ and angular speed $\omega$ is $v=r\omega$.

Conservation of energy yields that the decrease in gravitational potential or work done by gravitational force is equal to the gain in kinetic energy $\frac12mv^2+\frac12I\omega^2 = \frac12 m(1+k)v^2$.

If $k=1$ then the kinetic energy is equally shared by translation and rotation.
With smaller values of $k$ a larger proportion of the kinetic energy is with the translational motion and when $k=0$ there is no rotational kinetic energy and it is all translational kinetic energy.

Note that I wrote “gravitational potential or work done by gravitational force” which identifies clearly where the kinetic energy has come from.

The other way of doing this is consider the work done by the component of force parallel to the incline at angle $\alpha$ having moved a distance $s$ down the slope $(mg\sin \alpha-F_{\rm friction})s$ and the work done by the torque provided by the frictional force $F_{\rm friction}s$ which when added together gives $mgh$ where $h= s\sin \alpha$ is the vertical drop that the centre of mass of the object undergoes.

Update as a result of a comment from @An_African_Ape

$\tau = I \frac {d\omega}{dt} \Rightarrow F_{\rm friction} r \omega \,dt = I \omega \,d\omega \Rightarrow F_{\rm friction} r \,d\theta = I \omega \,d\omega$

Now do the integration $F_{\rm friction}r\theta = \frac 12 I \omega^2 \Rightarrow F_{\rm friction} \, s = \frac 12 I \omega^2$
So the work done by the torque is equal to the gain in rotational kinetic energy of the body.
A similar analysis can be done for the translational kinetic energy of the centre of mass of the body.

• The issue I'm having is that if you calculate the net force down the incline (gravity and friction) and multiply by the distance as you did at the end then that amount of work is not equal to the objects translational KE, and the same with the net torque multiplied by the angular distance not being equal to its rotational KE. Those calculations are independent of the objects moment of inertia but their rotational and translational KE is not. – An_African_Ape Sep 13 '18 at 13:15
• @An_African_Ape I have updated my answer with regard to the work done by the torque. – Farcher Sep 13 '18 at 15:59
• You'll have to forgive me here as I'm new to calculus and so I don't think I'm fully following. Since both objects in my example have the same torque through the same angular distance it seems as though they ought to have the same rotational KE but clearly they don't as I explained. Is there something about using calculus that fixes this problem? Wouldn't using your calculations give the same answer for the work done for both objects even though the one with the higher moment of inertia has more rotational KE? – An_African_Ape Sep 13 '18 at 17:58
• @An_African_Ape The frictional force does not have a constant value. It depends on the situation and would be zero for the point mass at the centre of a massless wheel. – Farcher Sep 13 '18 at 20:33

You say: "I think I'm just having a hard time getting my head around the fact that two objects can have the same net force acting ...".

They do not share the same net force. While $mgsin\theta$ is the same, static friction force is not. Static friction force can take any value (up to an upper limit) and is different for the two objects.

• I did consider this already but had a hard time understanding how a larger moment of inertia would result in a larger force of static friction. It also seems to me to suggest that if the force of static friction isn't always at its highest value for a rolling object then we can't know for sure (without doing the experiment) what the acceleration will be, only its maximum value. – An_African_Ape Sep 13 '18 at 18:52
• Your second issue is resolved mathematically: two equations (one translation and one rotation) and two unknowns (one friction force and one acceleration). Notice that linear acceleration determines angular acceleration. – npojo Sep 13 '18 at 19:11
• Assuming my calculations are correct I have been able to show that the force of static friction is proportional to K/1+K where "K" is the coefficient for the moment of inertia. This solves the issue I have been having. Thank you. – An_African_Ape Sep 14 '18 at 12:04

In the case of the rolling object you have to consider the work done on each individual volume element of it, because $\vec F \cdot d\vec s$ varies with time and position. Obviously the result will be different from that if the non rotating object.